# Martingale inequality

Let $f: \mathbb{R}_+ \times \mathbb{R}_+ \to \mathbb{R}$ be a deterministic function, as nice as you want, $W$ a Brownian motion and define $$Y^r_t := \int_0^t f(r,s) dW_s$$ For each fixed $r$, $(Y^r_t)_{t \geq 0}$ is a martingale and we can apply the following martingale inequality $$\mathbb{P} \left( \sup_{t \in (0,T]} |Y^r_t| \geq K_1, \langle Y^r_. \rangle _T \leq K_2\ \right) \leq 2 \exp \left( - \frac{K_1^2}{2K_2} \right)$$ The process I am really interested in is $\int_0^t f(t,s) dW_s$, i.e. $Y^t_t$. I would like to have a similar sort of inequality (and in fact it seems to be used in some papers that such an inequality holds) but $Y^t_t$ is not a martingale.

Can anyone explain why such an inequality would hold (e.g. if it is automatic from the above) or provide a reference or counterexample?

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Source?   – Did Apr 7 '13 at 12:25
@Did My source is a paper by Kusuoka & Stroock (Applications of the Malliavin calculus. II) which is not available online, but the inequality comes from viewing the martingale as Brownian Motion run at the "clock" of its quadratic variation, using the distribution the running max of a Brownian Motion and using Gaussian tail estimates. – mathman Apr 7 '13 at 12:43
@Did I have found a reference: see proposition 4.24 here statslab.cam.ac.uk/~beresty/teach/StoCal/sc3.pdf – mathman Apr 7 '13 at 12:54
Got it. As you explain, the trouble is that $(Y_t^t)_t$ is not guaranteed to be a martingale. Say, what makes you think such an extension holds? – Did Apr 7 '13 at 13:01
Well, the inequality holds for all fixed $r$. And, assuming $f$ is continuous in the first variable, I think that for a given $t$, $Y^r_t$ and $Y^t_t$ should be close when $r$ is close to $t$. However, I am not sure if the same is true of $\sup_{t \in [0,T]} Y^r_t$ and $\sup_{t \in [0,T]} Y^t_t$ or the quadratic variations. – mathman Apr 7 '13 at 13:31